Answer to Question #154864 in Differential Equations for (D^2 -4DD^'+ 3D^'^2)z =0

Question

(D^2 -4DD^'+ 3D^'^2)z =0

Answer 1

Rewrite this equation in the next view:

$z_{xx}-4z_{xy}+3z_{yy}=0$

Now lead it to canonical view. To do this, let input new variables:

$\begin{cases} \eta=y+x\\ \xi= y+3x\\ \end{cases}$

So now we have:

$\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial \eta}+3\dfrac{\partial z}{\partial \xi}$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \xi}$

$\dfrac{\partial^2 z}{\partial x^2}=(\dfrac{\partial z}{\partial \eta}+3\dfrac{\partial z}{\partial \xi})^2$

$\dfrac{\partial^2 z}{\partial y^2}=(\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \xi})^2$

$\dfrac{\partial^2 z}{\partial x \partial y}=\dfrac{\partial z^2}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \xi \partial\eta} + 9\dfrac{\partial z^2}{\partial \xi^2}$

$z_{\eta\eta}+6z_{\eta\xi}+9z_{\xi\xi}-4(z_{\eta\eta}+4z_{\eta\xi} +9z_{\xi\xi})+3(z_{\eta\eta}+2z_{\eta\xi}+z_{\xi\xi})=0$

The canonical equation:

$6z_{\xi\xi}+z_{\xi\eta}=0$

$\dfrac{\partial}{\partial \xi}(6\dfrac{\partial z}{\partial \xi} + \dfrac{\partial z}{\partial \eta})=0$

$6\dfrac{\partial z}{\partial \xi} + \dfrac{\partial z}{\partial \eta}=C_1(\eta)$

$\dfrac{\partial \xi}{6} =\partial \eta=\dfrac{\partial z}{C_1(\eta)}$

$z_1=\int C_1(\eta) d\eta +C_2$

$z_2=C_1(y+x)*(y+3x)+C_3$